Solve: bx+ay=a+b ax (1/a-b-1/a+b) +by (1/b-a …

The system of equation is given by :
bx + cy = a + b ......(i)
{tex}ax\left( {\frac{1}{{a - b}} - \frac{1}{{a + b}}} \right) + cy\left( {\frac{1}{{b - a}} + \frac{1}{{b + a}}} \right){/tex}{tex}= \frac{{2a}}{{a + b}}{/tex} ...(ii)
From equation (i)
bx + cy - (a + b) = 0 ............ (iii)
From equation (ii)
{tex} ax\left( {\frac{1}{{a - b}} - \frac{1}{{a + b}}} \right) + cy\left( {\frac{1}{{b - a}} + \frac{1}{{b + a}}} \right){/tex} {tex} - \frac{{2a}}{{a + b}} = 0{/tex}
{tex}⇒ x\left( {\frac{{2ab}}{{(a - b)(a + b)}}} \right) + y\left( {\frac{{2ac}}{{(b - a)(b + a)}}} \right){/tex} {tex}- \frac{{2a}}{{a + b}} = 0{/tex}
{tex} ⇒ \frac{1}{{a + b}}\left( {\frac{{2abx}}{{a - b}} - \frac{{2acy}}{{a - b}} - 2a} \right) = 0{/tex} 
{tex}⇒ \frac{{2abx}}{{a - b}} - \frac{{2acy}}{{a - b}} - 2a = 0{/tex}
  2abx - 2acy - 2a(a - b) = 0 ....(iv)
From equation (iii) and (iv), we get
a₁ = b, b₁ = c and c₁ = - (a + b)
and a₂ =  2ab , b₂ = -2ac  and c₃ = -2a(a - b)
by cross-multiplication, we get
{tex}\frac{x}{{ - 4{a^2}c}} = \frac{{ - y}}{{4a{b^2}}} = \frac{{ - 1}}{{4abc}}{/tex}
Now, {tex}\frac{x}{{ - 4{a^2}c}} = \frac{{ - 1}}{{4abc}} {/tex}
{tex}⇒ x = \frac{a}{b}{/tex}
And, {tex}\frac{{ - y}}{{4a{b^2}}} = \frac{{ - 1}}{{4abc}} {/tex}
{tex}⇒ y = \frac{b}{c}{/tex}
The solution of the system of equation are {tex}\frac{a}{b}{/tex} and {tex}\frac{b}{c}{/tex}.